Bloch's theorem and Bloch's state

Question

The question is not so much about the theorem, but more about what it means in this context: see this link.

So yes, because of Bloch's theorem the Hamiltonian eigenstates in a crystalline system can be written as \begin{align} \psi_{n,\vec{k}}(\vec{r})=e^{i\vec{k}\cdot\vec{r}}u_{n,\vec{k}}(\vec{r}), \end{align} and so the Berry connection can be defined: \begin{align} A_{n}(\vec{k})=i\langle n(\vec{k})|\nabla_{\vec{k}}|n(\vec{k})\rangle, \end{align} but what in the world is $|n(\vec{k})\rangle$?

I've read a few articles on topological insulators and they always seem to start off with the Bloch wavefunction $e^{i\vec{k}\cdot\vec{r}} u_k(\vec{r})$, and then somehow they magically get the ket $|u(\vec{k})\rangle$ from which the Berry connection is defined... is $|u(\vec{k})\rangle$ the column vector comprised of the Fourier coefficients of $u_\vec{k}(\vec{r})$ w.r.t. $e^{i\vec{G}\cdot\vec{r}}$ or what?

This post imported from StackExchange Physics at 2014-03-24 04:14 (UCT), posted by SE-user nervxxx

Comments

As the article says, $n$ is the band index. If you were to represent the ket $|u_{n}(\mathbf{k})\rangle$ in a basis (in this case the basis of a Hilbert space spanned by bands) then you could write it as a column vector where each component corresponds to the Bloch wavefunction for each band (labeled by $n$). Note: for a the simplest topological insulator model you need at least two bands: valence and conduction band. In that case you'll have a $2 \times 1$ column vector.

This post imported from StackExchange Physics at 2014-03-24 04:14 (UCT), posted by SE-user NanoPhys

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@NanoPhys How do you get rid of the position $\vec{r}$ dependence then? Is that absorbed in the way the inner product is defined? But $|n(\vec{k})\rangle$ should exist on its own, and should be independent of $\vec{r}$. How does one get it from $u_{n,\vec{k}}(\vec{r})$?

This post imported from StackExchange Physics at 2014-03-24 04:14 (UCT), posted by SE-user nervxxx

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Yes, that is correct. The $\mathbf{r}$ does indeed get absorbed in the definition of the inner product. Note that a Bloch state is uniquely labeled by $n$ and $\mathbf{k}$ independent of what basis it is represented in. In your case, you are writing down the Bloch "wavefunction" $\psi_{n,\mathbf{k}}(\mathbf{r})\propto e^{i\mathbf{k}\cdot\mathbf{r}}u_{n,\mathbf{k}}(\mathbf{r})$ in the position basis. The inner product $\langle u_{n}(\mathbf{k})|\dots|u_{n}(\mathbf{k})\rangle$ has to be basis independent.

This post imported from StackExchange Physics at 2014-03-24 04:14 (UCT), posted by SE-user NanoPhys

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@NanoPhys So what would Bloch's theorem be, without going into any explicit basis (in particular, the position basis)? What I mean is, what can we say about the abstract kets $|\psi\rangle$? See arxiv.org/pdf/1304.5693v3.pdf , eqn 27-30. I'm not sure why that is true unless he's already assuming that the kets are already in the position basis

This post imported from StackExchange Physics at 2014-03-24 04:14 (UCT), posted by SE-user nervxxx

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Equation (28) is valid. The $e^{i\mathbf{k}\cdot\mathbf{r}}$ is simply a phase factor; two kets can, in general, be related in that fashion. To make better sense of it, take a $\langle\mathbf{r}|$ on both sides. You'll get $\langle\mathbf{r}|\psi_{n\mathbf{k}}\rangle = e^{i \mathbf{k} \cdot\mathbf{r}} \langle \mathbf{r}|u_{n\mathbf{k}}\rangle\Rightarrow\psi_{n\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}}u_{n\mathbf{k}}(\mathbf{r})$, which is our familiar Bloch wavefunction

This post imported from StackExchange Physics at 2014-03-24 04:14 (UCT), posted by SE-user NanoPhys